# partial differentiation on differentials

I get the idea behind partial differentiation but this one is really tricky!

If $z = xe^{-y}$, and $x = \cosh t$, and $y = \cos s$, then what is the partial of $z$ with respect to $s$, and partial of $z$ with respect to $t$?

Thanks

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There are two ways to do this:

1.) just plug in the expressions for $x$ and $y$ in terms of $s$ and $t$ and differentiate.

2.) use the multivariate chain rules:

$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$$ $$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$$

Make sense?

By request, $$dz = d(xe^{-y}) =e^{-y}dx+-xe^{-y}dy$$ however, since $x = \cosh(t)$ and $y = \cos(s)$ we find $$dx = \sinh(t)dt \qquad \text{and} \qquad dy = -\sin(s)ds$$ plug those into the $dz$ formula to obtain: $$dz = e^{-y}\sinh(t)dt -xe^{-y}(-\sin(s)ds)$$ We can read from the above the coefficient of $dt$ is $\frac{\partial z}{\partial t}$ and the coefficient of $ds$ is $\frac{\partial z}{\partial s}$. Of course I use $x$ and $y$ here as abbreviations for the $t,s$ formulas. Intuitively, the partial derivative w.r.t. $t$ is when $s$ is held constant so $ds=0$ and this is why this approach works. In advanced calculus we can give better answers in terms of the implicit or inverse function theorems. As a general principle, you can use differentials and proceed formally, this approach goes a long way.

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If you want to know why these are the correct formulas I can elaborate. –  James S. Cook Sep 15 '12 at 2:28
It makes sense and partial of x w.r.t. s = 0 and partial of y w.r.t. t = 0 –  mary Sep 15 '12 at 21:32
@mary correct, I'm just giving the answer that works no matter what the formulas you get for x and y. You said "differentials" in the original post, did you want to see more about that? I known a way to do all of this formally by pushing differentials around formally.... if you're interested. –  James S. Cook Sep 16 '12 at 2:27
Please go ahead and show me. Thans –  mary Sep 17 '12 at 4:36
@MykeArya not really, I understood the method of differentials far before I understood the exterior calculus. Moreover, I also gave your answer in my original post, granted you have included a few details which are nice. –  James S. Cook Apr 12 '13 at 17:42
$\frac{\partial{z(x,y)}}{\partial{s}}=\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}+\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{s}}=\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}+0=\{-xe^{-y}\}\{-\sin(s)\}$
$\frac{\partial{z(x,y)}}{\partial{t}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{t}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{t}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{t}}+0=\{e^{-y}\}\{-\sinh(s)\}$
It's just applying chain rule and noticing that $\frac{\partial{x}}{\partial{s}}=0$ because $x$ does not depend on $s$. Therfore is constant regarding this variable, and the derivative of a constant vanishes. And similarty $\frac{\partial{y}}{\partial{t}}=0$.